Tensor in different coordinate system I have the tensors $F_{\mu\nu}$, $F^{\mu\nu}$ in coordinate system $(t,x,y,z)$ and want to transform these to coordinate system $(t',x',y',z')$ just by multiplicating matrices.  
My idea was to calculate the Jacobians $J=(\frac{\partial x^i}{\partial x'^j})_{ij}$ and $J'=(\frac{\partial x'^i}{\partial x^j})_{ij}$.
Then I would find $$F'_{\mu\nu}=J^\top F_{\mu\nu}J$$ and $$F'^{\mu\nu}=J' F^{\mu\nu}J'^\top,$$ in matrix notation.

Is this correct?

My ultimate goal is to prove that $F_{\mu\nu}F^{\mu\nu}$ is the same in both systems, however calculating this explicitly does not give me this result.
 A: The idea is more or less correct but you need to be careful with the mixing of matrix and tensor notation. Let's say that $F_{\mu\nu}$ are the components of a matrix $F$, and $F^{\mu\nu}$ those of $F_U$ (for "upper"). Then $F_{\mu\nu}F^{\mu\nu}$ becomes $\operatorname{tr}(F F_U^T)$. Also you need to notice that $J'$ is the inverse of $J$.
With this notation, the transformations are $F' = J^T F J$ and $F_U' = J' F_U J'^T$, and the trace is transformed to $\operatorname{tr}(J^T F J J' F_U^T J'^T) = \operatorname{tr}(J^T F F_U^T J'^T) = \operatorname{tr}(F F_U^T J'^T J^T) = \operatorname{tr}(F F_U^T)$, using the cyclic property of the trace in the middle.
A: By definition, under a change in coordinate system a tensor's components transform(s)
$$F_{\mu'\nu'}=\frac{\partial x^{\mu'}}{\partial x^{\mu}}\frac{\partial x^{\nu'}}{\partial x^{\nu}}F_{\mu'\nu'}$$
So you could prove your statement by showing
$$F_{\mu'\nu'}F^{\mu'\nu'}=F_{\mu\nu}F^{\mu\nu}$$
by using the metric $\eta_{\mu\nu}$ to raise and lower the indices on the definition of the transformation.
